Question

let θ be an angle such that cotθ = -8/15 and sinθ < 0. find the exact values of cscθ and secθ. cscθ = secθ =

Explanation:

Step1: Recall the relationship between cotangent and sine/cosine

We know that $\cot\theta=\frac{\cos\theta}{\sin\theta}=-\frac{8}{15}$. Let $\cos\theta = - 8k$ and $\sin\theta=15k$ for some non - zero real number $k$.

Step2: Use the Pythagorean identity $\sin^{2}\theta+\cos^{2}\theta = 1$

Substitute $\cos\theta=-8k$ and $\sin\theta = 15k$ into the identity: $(15k)^{2}+(-8k)^{2}=1$.
Expanding gives $225k^{2}+64k^{2}=1$, so $289k^{2}=1$. Then $k^{2}=\frac{1}{289}$, and $k=\pm\frac{1}{17}$.

Step3: Determine the sign of $k$

Since $\sin\theta<0$, and $\sin\theta = 15k$, then $k<0$. So $k =-\frac{1}{17}$.

Step4: Find the value of $\sin\theta$

$\sin\theta=15k=15\times(-\frac{1}{17})=-\frac{15}{17}$.
Since $\csc\theta=\frac{1}{\sin\theta}$, then $\csc\theta=-\frac{17}{15}$.

Step5: Find the value of $\cos\theta$

$\cos\theta=-8k=-8\times(-\frac{1}{17})=\frac{8}{17}$.
Since $\sec\theta=\frac{1}{\cos\theta}$, then $\sec\theta=\frac{17}{8}$.

Answer:

$\csc\theta=-\frac{17}{15}$
$\sec\theta=\frac{17}{8}$